How do you simplify root(3)(135+78sqrt(3))+root(3)(135-78sqrt(3)) ?

2 Answers
Apr 20, 2018

root(3)(135+78sqrt(3))+root(3)(135-78sqrt(3)) = 6

Explanation:

Let:

x = root(3)(135+78sqrt(3))+root(3)(135-78sqrt(3))

Then:

x^3 = (135+78sqrt(3))+(135-78sqrt(3))+3(root(3)(135+78sqrt(3))root(3)(135-78sqrt(3)))(root(3)(135+78sqrt(3))+root(3)(135-78sqrt(3)))

color(white)(x^3) = 270+3(root(3)(135^2-78^2(3)))x

color(white)(x^3) = 270+3(root(3)(18225-18252))x

color(white)(x^3) = 270+3(root(3)(-27))x

color(white)(x^3) = 270-9x

So:

x^3+9x-270 = 0

By the rational roots theorem, any rational roots of this cubic are expressible in the form p/q for integers p, q with p a divisor of the constant term -270 and q a divisor of the coefficient 1 of the leading term.

So the first few possible rational zeros are:

+-1, +-2, +-3, +-5, +-6, +-9, +-10,...

By Descartes' Rule of Signs, we can tell that the cubic has exactly one positive real root and no negative real zeros, so try the possible positive rational zeros.

We find:

(color(blue)(6))^3+9(color(blue)(6))-270 = 216+54-270 = 0

So x=6 is a root and (x-6) is a factor:

x^3+9x-270 = (x-6)(x^2+6x+45)

The remaining quadratic has no real zeros (as we can see by looking at its discriminant).

The non-real zeros are actually:

omega root(3)(135+78sqrt(3))+omega^2 root(3)(135-78sqrt(3))

and:

omega^2 root(3)(135+78sqrt(3))+omega root(3)(135-78sqrt(3))

where omega = -1/2+sqrt(3)/2i is the primitive complex cube root of 1.

So:

root(3)(135+78sqrt(3))+root(3)(135-78sqrt(3)) = 6

Apr 20, 2018

6.

Explanation:

Let us note that,

(x+sqrt3y)^3=x^3+3sqrt3y^3+3sqrt3xy(x+sqrt3y),

=x^3+3sqrt3y^3+3sqrt3x^2y+9xy^2.

rArr (x+sqrt3y)^3=(x^3+9xy^2)+3sqrt3(x^2y+y^3), or,

root(3){(x^3+9xy^2)+sqrt3(3x^2y+3y^3)}=x+sqrt3y.

So, from this we guess :

If root(3)(135+78sqrt3)=x+sqrt3y, then,

(x^3+9xy^2)=135, and, (3x^2y+3y^3)=78, i.e.,

x^3+9xy^2=x(x^2+9y^2)=135......(1), and,

3y(x^2+y^2)=78, or, y(x^2+y^2)=26......(2).

y(x^2+y^2)=26=1xx26, or, 2xx13

By Trial and Error Method, y=1, x=5;" not suitable for "(1)

y=2, x=3" satisfy "(1).

Therefore, root(3)(135+78sqrt3)=x+sqrt3y=3+2sqrt3.

Consequently, root(3)(135-78sqrt3)=3-2sqrt3.

Finally, root(3)(135+78sqrt3)+root(3)(135-78sqrt3)=6,

as Respected George C. Sir has already derived!